3243 - FlattenDepth

3243 - FlattenDepth

by jiangshan (@jiangshanmeta) #中等 #array

题目

Recursively flatten array up to depth times.

For example:

type a = FlattenDepth<[1, 2, [3, 4], [[[5]]]], 2> // [1, 2, 3, 4, [5]]. flattern 2 times
type b = FlattenDepth<[1, 2, [3, 4], [[[5]]]]> // [1, 2, 3, 4, [[5]]]. Depth defaults to be 1

If the depth is provided, it's guaranteed to be positive integer.

在 Github 上查看：https://tsch.js.org/3243/zh-CN

代码

/* _____________ 你的代码 _____________ */

type FlattenDepth<T extends any[], U extends number = 1, C extends any[] = []> =
  // Check if the current depth reaches the specified maximum depth
  C['length'] extends U
    ? T
    : T extends [infer F, ...infer R]
      // Check if the first element is an array
      ? F extends any[]
        // Recursively flatten the first element and increment the depth
        ? [...FlattenDepth<F, U, [...C, 1]>, ...FlattenDepth<R, U, C>]
        : [F, ...FlattenDepth<R, U, C>]
      : T

测试用例

/* _____________ 测试用例 _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<FlattenDepth<[]>, []>>,
  Expect<Equal<FlattenDepth<[1, 2, 3, 4]>, [1, 2, 3, 4]>>,
  Expect<Equal<FlattenDepth<[1, [2]]>, [1, 2]>>,
  Expect<Equal<FlattenDepth<[1, 2, [3, 4], [[[5]]]], 2>, [1, 2, 3, 4, [5]]>>,
  Expect<Equal<FlattenDepth<[1, 2, [3, 4], [[[5]]]]>, [1, 2, 3, 4, [[5]]]>>,
  Expect<Equal<FlattenDepth<[1, [2, [3, [4, [5]]]]], 3>, [1, 2, 3, 4, [5]]>>,
  Expect<Equal<FlattenDepth<[1, [2, [3, [4, [5]]]]], 19260817>, [1, 2, 3, 4, 5]>>,
]

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